Answer:
temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L
Explanation:
property table for R-134a
https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html
at 160 KPa , temperature = -15.66 C
quality x=mass of vapour/ total mass of liq-vap mixture
alternaternately: x=(v-vf)/(vg-vf)
v=total volume i.e. volume of container"80L" 80L=0.08 cubic meter
vf=vol of liquid phase vg=vol of vapor phase vf, vg values at 160Kpa
x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646
enthalpy
h=hf+xhfg hf, hfg values at 160Kpa
h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg
for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ
volume of vapor phase
vg at 160Kpa=0.1235 cubic meter for quality=1.
in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.
vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L
